Can someone help me put this Ellipse Equation into standard form? (10 pts!)?

Question

Here is what standard form looks like for ellipses:
[(x - h)^2 / a^2] + [(y - k)^2 / b^2] = 1

or
[(x - h)^2 / b^2] + [(y - k)^2 / a^2] = 1


(h,k) is center, a is the distance from center to vertices, b is the distance from center to co-vertices.

Here is the equation I want to put into standard form:
3(x^2 + y^2) - 8x - 4y - 4 = 0

Answer 1

Start with:

3(x² + y²) - 8x - 4y - 4 = 0;

Expanding:

3x² + 3y² -8x -4y -4 = 0;

Regrouping:

3x²-8x +3y² -4y = 4;

Dividing by 3:

x² - (8/3)x + y² - (4/3)y = 4/3

Okay, here's the part that's hard to see: you have two pieces (one in x and one in y) of the left-hand side that look like the pieces of a quadratic expression. What we're going to do is complete the square of each of the pieces and factor them as a perfect square. So for example:

x² - (8/3)x = x² - (8/3)x + (16/9) - 16/9
= (x-4/3)² - 16/9

You complete the square by dividing the coefficient in front of x to the first power by 2, then squaring the result. So 8/3 -> 4/3 -> 16/9. Got it?

Similarly, for the y terms:

y² - (4/3)y = y² - (4/3)y +4/9 - 4/9
= (y-2/3)² -4/9

So resuming from x² - (8/3)x + y² - (4/3)y = 4/3, we substitute the perfect-square expressions to get:

(x-4/3)² - 16/9 + (y-2/3)² -4/9 = 4/3 = 12/9
(x-4/3)² + (y-2/3)² = 12/9 + 16/9 + 4/9 = 32/9

Divide both sides by 9/32 to get:

(9/32)(x-4/3)² + (9/32)(y-2/3)² = 1

So mapping the terms to your standard form,
h=4/3
k=2/3
a=√(32/9) = (4/3)√2
b=√(32/9) = (4/3)√2

So your ellipse is actually a circle of radius (4/3)√2, with a center at x=4/3, y=2/3. And yes, all circles are ellipses.

Good luck, work hard, and stay away from drugs.

Answer 2

you must complete the square for both x and y 4x^2 - 8x ....+y^2 + 3y ..... = 19 4(x^2 - 2x ......) + y^2 + 3y .......=19 find the number needed = (b/2)^2 FOR x need +1 for y need +9/4... balance equation when putting in blamks 4(x^2-2x+ 1) + y^2 + 3y + 9/4 = 19 ( add to both sides)+4+9/4 4(x-1)^2 + (y+3/2)^2 = 19+4+(9/4) THIS IS NOT A CIRCLE!!!!! It is an ellipse! Divide by 4 (x-1)^2 + (y+3/2)^2 / 4 = 101/16 OR standard form for ellipse: (x-1)^2 + (y+3/2)^2 = 101 ____________________ . . . . . . . . . 2^2..........4^2

Answer 3

Let's divide out the 3 first.
x^2 + y^2 -8/3 x -4/3y - 4/3= 0
What you need to do next is form two perfect squares (x - 4/3)^2 + (y - 2/3)^2 = 0
Hopefully, this will really work and the constant will be swallowed up in the perfect square formation.

Answer 4

group the x and y together and then complete the square.

for example: 3(x^2-8/3x+ 16/9 )-3y^2-4y-4 = 16/3

eventually if you keep solving you'll get the standard equation.

Answer 5

I'm not a math nerd but it's really just combining like terms.
3(x^2+y^2)-8x-4y-4=0
3x^6+3y^6-8x-4y-4=0
-5x^2-1y^6=0

I don't know, something like that. I'd check out a math book or something at a library.

Answer 6

Complete square

3(x^2-8/3x)+3(y^2-4/3y)=4

3(x^2-8/3x+(4/3)^2)+3(y^2-4/3y+(2/3)^2)=4

3(x-4/3)^2+3(y-2/3)^2=4+20/3

(x-4/3)^2+(y-2/3)^2=32/9

(x-4/3)^2/(32/9) +(y-2/3)^2/(32/9) =1

Please check the operations.